Iincluded a photo with the information! here are the questions. a thorough explanation and diagram would immensely. you so much!

question 1: reate a labeled diagram of the brick. use x to represent the length of a side of the base of the candle. then write an equation for the volume of the candle in terms of x. write the polynomial in standard form.

question 2: create a labeled diagram of the egyptian. use x to represent the length of a side of the base of the candle. then write an equation for the volume of the candle in terms of x. write the polynomial in standard form.

question 3: 1. to impress management, you decide to propose your own original candle design based on solid figures. make your candle design attractive and make sure it will be practical to manufacture. draw the candle in the space below and label the drawing with dimensions. then write an equation for the volume of the candle in terms of x, where x is a dimension that defines the area of the candle’s base. write the polynomial in standard form.

1) Labeled diagram in the first figure attached

Volume of the rectangular prism: x^2*(x-1) = x^3 - x^2 ; x in cm, volume in cm^3

2) Labeled diagram in the second figure attached

Volume of the right square pyramid: (1/3)*(x^2)*21 = 7*x^2, x in cm, volume in cm^3

3) A cylinder is proposed, where x is the radius of the base and the height is 4 times the radius. Labeled diagram in the third figure attached .

Volume of the cylinder: π*x^2*(4*x) = 4*π*x^3, x in cm, volume in cm^3

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3